How to create complex numpy array by skipping each column one by one

Question

I have an complex (3x6) numpy array like this:

x = [[(1+j2) ,(2+j8), (4+j1), (6+j8), (7+j3), (8+j2)],
     [(3+j8), (5+j1), (7+j5), (3+j2), (6+j1), (3+j1)],
     [(1+j5), (5+j4), (2+j9), (9+j5), (8+j1), (4+j1)]]

I want to split them as:

x1 = [[(1+j2), (4+j1), (7+j3)],
      [(3+j8), (7+j5), (6+j1)],
      [(1+j5), (2+j9), (8+j1)]]

and

x2 = [[(2+j8), (6+j8), (8+j2)],
      [(5+j1), (3+j2), (3+j1)),
      [(5+j4), (9+j5), (4+j1)]]

I mean, I want to get a new complex numpy array by skipping each column one by one. How can I do it?

Matteo Zanoni · Accepted Answer · 2021-05-05 16:07:24Z

2

This can be achieved using array indexing.

import numpy as np

x = np.array(
    [
        [(1 + 2j), (2 + 8j), (4 + 1j), (6 + 8j), (7 + 3j), (8 + 2j)],
        [(3 + 8j), (5 + 1j), (7 + 5j), (3 + 2j), (6 + 1j), (3 + 1j)],
        [(1 + 5j), (5 + 4j), (2 + 9j), (9 + 5j), (8 + 1j), (4 + 1j)],
    ]
)
x1 = x[:, ::2]
x2 = x[:, 1::2]

Yu can find a more detailed explanation of array indexing here

answered May 5, 2021 at 16:07

Matteo Zanoni

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Comments

Matheus Muriel · Accepted Answer · 2021-05-05 16:11:22Z

0

For this you can use list comprehension

Exemple:

x = [[(1+j2) ,(2+j8), (4+j1), (6+j8), (7+j3), (8+j2)],
     [(3+j8), (5+j1), (7+j5), (3+j2), (6+j1), (3+j1)],
     [(1+j5), (5+j4), (2+j9), (9+j5), (8+j1), (4+j1)]]

x1 = [ i[0:6:2] for i in x ]
x2 = [ i[1:6:2] for i in x ]

answered May 5, 2021 at 16:11

Matheus Muriel

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